The Statement of the Problem:
Let $ \{ N(t), t \ge 0 \} $ be a Poisson process with $\lambda = 1$. Define
$$A(t) = \begin{cases} t & \text{if $N(t) = 0$} \\ t-s_k & \text{if $N(t) = k, k \ge 1$.} \end{cases}$$
Find the distribution of $A(t)$, then find the mean of $A(t)$.
SOLUTION
$$ \mathbb P \{A(t)=0 \} = \mathbb P \{N(t)=0 \} = e^{-t} $$
and for $0 \lt x \lt t$,
$$ \mathbb P \{A(t) \gt x \} = \mathbb P \{N(t) - N(t-x) =0 \} = e^{-x}. $$
So the distribution of $A(t)$ is truncated exponential; that is,
$$ A(t) \sim \min\{Y, t\} $$
where $Y \sim \text{Expo}(1)$ and
$$ \mathbb P \{A(t) \gt x \} = \begin{cases} e^{-x} & \text{if $0 \le x \lt t$} \\ 0 & \text{if $ x \ge t$.} \end{cases} $$
Now, the mean of $A(t)$ is
$$ \mathbb E \{A(t) \} = \int_{0}^{t} \mathbb P \{A(t) \gt x \}dx = \int_{0}^{t} e^{-x}dx = 1 - e^{-t}.$$
My Question:
Why isn't the mean
$$ \mathbb E \{A(t) \} = \int_{0}^{t}x \mathbb P \{A(t) \gt x \}dx = \int_{0}^{t} xe^{-x}dx?$$
As in, why isn't it computed the usual way we compute expectation by throwing an additional "$x$" in the integrand with the pdf?